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Sum of integers upto N with given unit digit (Set 2)
  • Difficulty Level : Medium
  • Last Updated : 20 Apr, 2021

Given two integer N and D where 1 ≤ N ≤ 1018, the task is to find the sum of all the integers from 1 to N whose unit digit is D.
Examples: 
 

Input: N = 30, D = 3 
Output: 39 
3 + 13 + 23 = 39
Input: N = 5, D = 7 
Output:
 

 

Approach: In Set 1 we saw two basic approaches to find the required sum, but the complexity is O(N) which will take more time for larger N. Here’s an even efficient approach, suppose we are given N = 30 and D = 3
 

sum = 3 + 13 + 23 
sum = 3 + (10 + 3) + (20 + 3) 
sum = 3 * (3) + (10 + 20) 
 



From the above observation, we can find the sum following the steps below: 
 

  • Decrement N until N % 10 != D.
  • Find K = N / 10.
  • Now, sum = (K + 1) * D + (((K * 10) + (10 * K * K)) / 2).

Below is the implementation of the above approach: 
 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define ll long long int
 
// Function to return the required sum
ll getSum(ll n, int d)
{
    if (n < d)
        return 0;
 
    // Decrement N
    while (n % 10 != d)
        n--;
 
    ll k = n / 10;
 
    return (k + 1) * d + (k * 10 + 10 * k * k) / 2;
}
 
// Driver code
int main()
{
    ll n = 30;
    int d = 3;
    cout << getSum(n, d);
    return 0;
}

Java




// Java  implementation of the approach
 
import java.io.*;
 
class GFG {
 
 
// Function to return the required sum
static long getSum(long n, int d)
{
    if (n < d)
        return 0;
 
    // Decrement N
    while (n % 10 != d)
        n--;
 
    long k = n / 10;
 
    return (k + 1) * d + (k * 10 + 10 * k * k) / 2;
}
 
// Driver code
 
    public static void main (String[] args) {
     long n = 30;
    int d = 3;
    System.out.println(getSum(n, d));    }
}
//This code is contributed by inder_verma..

Python3




# Python3 implementation of the approach
 
# Function to return the required sum
def getSum(n, d) :
     
    if (n < d) :
        return 0
 
    # Decrement N
    while (n % 10 != d) :
        n -= 1
 
    k = n // 10
 
    return ((k + 1) * d +
            (k * 10 + 10 * k * k) // 2)
 
# Driver code
if __name__ == "__main__" :
 
    n = 30
    d = 3
    print(getSum(n, d))
 
# This code is contributed by Ryuga

C#




// C# implementation of the approach
 
 
class GFG {
 
 
// Function to return the required sum
static int getSum(int n, int d)
{
    if (n < d)
        return 0;
 
    // Decrement N
    while (n % 10 != d)
        n--;
 
    int k = n / 10;
 
    return (k + 1) * d + (k * 10 + 10 * k * k) / 2;
}
 
// Driver code
 
    public static void Main () {
    int n = 30;
    int d = 3;
    System.Console.WriteLine(getSum(n, d)); }
}
//This code is contributed by mits.

PHP




<?php
// PHP implementation of the approach
 
// Function to return the required sum
function getSum($n, $d)
{
    if ($n < $d)
        return 0;
 
    // Decrement N
    while ($n % 10 != $d)
        $n--;
 
    $k = (int)($n / 10);
 
    return ($k + 1) * $d +
           ($k * 10 + 10 * $k * $k) / 2;
}
 
// Driver code
$n = 30;
$d = 3;
echo getSum($n, $d);
 
// This code is contributed by mits
?>

Javascript




<script>
 
// java script  implementation of the approach
 
// Function to return the required sum
function getSum(n, d)
{
    if (n < d)
        return 0;
 
    // Decrement N
    while (n % 10 != d)
        n--;
 
    k = parseInt(n / 10);
 
    return (k + 1) * d +
        (k * 10 + 10 * k * k) / 2;
}
 
// Driver code
let n = 30;
let d = 3;
document.write( getSum(n, d));
 
// This code is contributed
// by bobby
</script>
Output: 
39

 

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